A thin rod of length L is lying along the x-axis with its ends at x = 0 and x = L. Its linear densit

Q: Why do we use integration to find the center of mass here?

Because the mass is not concentrated at discrete points but is spread continuously along the rod with a varying density. Integration allows us to sum up the infinite number of infinitesimally small mass elements.

Q: What does $n=0$ physically mean in this problem?

If $n=0$, the term $(x/L)^0$ becomes $1$, making the density $\lambda(x) = k$, which is a constant. This represents a uniform rod, and its center of mass is naturally at the geometric center, $L/2$.

Q: Why does the center of mass approach $L$ as $n$ approaches infinity?

As $n$ gets larger, the fraction $(x/L)^n$ becomes extremely small for $x < L$ and only becomes significant very close to $x = L$. This means almost all the mass is concentrated at the rightmost tip of the rod.

Q: How do we know the curve is concave down?

If you take the first derivative of $x_{CM}$ with respect to $n$, it is positive (the function is increasing). The second derivative is negative, which mathematically proves the curve is concave down, meaning it flattens out as it approaches $L$.

Q: Can the center of mass ever be exactly at $L$?

Mathematically, it only reaches $L$ in the limit as $n \rightarrow \infty$. For any finite value of $n$, there is still some mass distributed along the rod, so the center of mass will always be slightly less than $L$.

Question

A thin rod of length $L$ is lying along the x-axis with its ends at $x = 0$ and $x = L$. Its linear density (mass/length) varies with $x$ as $k \left(\frac{x}{L}\right)^n$, where $n$ can be zero or any positive number. If the position $x_{CM}$ of the centre of mass of the rod is plotted against $n$, which of the following graphs best approximates the dependence of $x_{CM}$ on $n$ ?

Options:

(A)

(B)

(C)

(D)

Sigma Percentile · Accepted Answer

### The Setup: A Rod with a Twist

Imagine a thin rod lying along the $x$-axis, stretching from $x = 0$ to $x = L$. In a perfect, uniform world, its center of mass would sit exactly in the middle at $L/2$. But this rod has a twist: its mass is not distributed evenly. The linear density, which is the mass per unit length, is given by the function $\lambda(x) = k \left(\frac{x}{L}ight)^n$.

Notice that as $x$ increases from $0$ to $L$, the density increases. This means the rod gets heavier towards the right end. The parameter $n$ controls exactly *how* drastically this density changes. Our goal is to figure out how the position of the center of mass, $x_{CM}$, shifts as we change this power $n$.

### The Master Equation: Center of Mass

To find the center of mass of a continuous body, we rely on the fundamental integral formula. The position $x_{CM}$ is the mass-weighted average of all the positions $x$ along the rod:

$$x_{CM} = \frac{\int x dm}{\int dm}$$

Here, the infinitesimally small mass element $dm$ is simply the linear density $\lambda(x)$ multiplied by the small length element $dx$. Substituting our density function, the equation becomes:

$$x_{CM} = \frac{\int_0^L x \cdot k \left(\frac{x}{L}ight)^n dx}{\int_0^L k \left(\frac{x}{L}ight)^n dx}$$

### The Calculus: Integrating the Density

Let's tackle the numerator and denominator separately. The constants $k$ and $L^n$ can be pulled out of the integrals.

**For the numerator:**

$$	ext{Numerator} = \frac{k}{L^n} \int_0^L x^{n+1} dx$$

Using the basic power rule of integration, we get:

$$	ext{Numerator} = \frac{k}{L^n} \left[ \frac{x^{n+2}}{n+2} ight]_0^L = \frac{k L^2}{n+2}$$

**For the denominator (which is the total mass):**

$$	ext{Denominator} = \frac{k}{L^n} \int_0^L x^n dx$$

Again, applying the power rule:

$$	ext{Denominator} = \frac{k}{L^n} \left[ \frac{x^{n+1}}{n+1} ight]_0^L = \frac{k L}{n+1}$$

### The Grand Reveal: Analyzing the Result

Now, we divide the numerator by the denominator to find $x_{CM}$:

$$x_{CM} = \frac{\frac{k L^2}{n+2}}{\frac{k L}{n+1}} = L \left( \frac{n+1}{n+2} ight)$$

This elegant little formula tells us exactly where the center of mass is for any value of $n$. Let's test it at the extremes to understand its behavior.

**Case 1: $n = 0$**
If $n = 0$, the density $\lambda(x) = k$, which means the rod is uniform. Plugging $n=0$ into our formula gives $x_{CM} = L \left(\frac{1}{2}ight) = \frac{L}{2}$. This makes perfect physical sense!

**Case 2: $n ightarrow \infty$**
As $n$ becomes infinitely large, the term $\frac{n+1}{n+2}$ approaches $1$. Therefore, $x_{CM} ightarrow L$. Physically, a massive $n$ means almost all the mass is concentrated at the very tip of the right end, pulling the center of mass all the way to $L$.

### The Graphical Interpretation

We need to plot $x_{CM}$ against $n$. We know the curve starts at $(0, L/2)$ and asymptotically approaches the horizontal line $x_{CM} = L$.

To determine the shape of the curve, we can rewrite the function slightly:

$$x_{CM}(n) = L \left( 1 - \frac{1}{n+2} ight)$$

As $n$ increases, the subtracted term $\frac{1}{n+2}$ gets smaller, meaning $x_{CM}$ is strictly increasing. Furthermore, the rate at which it increases slows down as $n$ gets larger. Mathematically, the second derivative is negative, which means the curve is **concave down**.

Looking at the options provided, the graph that starts at $L/2$, increases, and curves downwards to approach $L$ asymptotically is the correct choice.

Select Answer:

Visualized Solution

$The Non-Uniform Rod$

$Center of Mass Formula$

$Substituting Density$

$Evaluating Numerator$

$Evaluating Denominator$

$Position of Center of Mass$

$Analyzing the Function$

$Graph of x_{C M} vs n$

$The Way Forward$

The Sigma Insight: Centre of Mass

The Setup

The Master Equation

The Calculus

The Grand Reveal

The Graphical Interpretation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Non-Uniform Rod

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Center of Mass Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Density

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Numerator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluating Denominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of Center of Mass

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Function

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Graph of xCM​ vs n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward